1.2题，求解
答案:3 悬赏:0 手机版
解决时间 2021-01-31 08:56
- 提问者网友：末路
- 2021-01-30 12:03
1.2题，求解
最佳答案
- 五星知识达人网友：蓝房子
- 2021-01-30 13:41
1、令x=tant，dx=sec^2tdt
原式=2*∫(tant+1)/[(tant-1)sec^4t]*sec^2tdt
=2*∫(sint+cost)/(sint-cost)*cos^2tdt
=∫(sint+cost)^2/(sin^2t-cos^2t)*(1+cos2t)dt
=∫(1+sin2t)/(-cos2t)*(1+cos2t)dt
=-∫(1+sin2t)/cos2tdt-∫(1+sin2t)dt
=-∫sec2tdt-∫tan2tdt-∫(1+sin2t)dt
=-(1/2)*ln|sec2t+tan2t|-(1/2)*ln|sec2t|-t+(1/2)*cos2t+C
=(1/2)*[cos2t-2t-ln|sec2t+tan2t|-ln|sec2t|]+C
=(1/2)*[(1-x^2)/(x^2+1)-2arctanx-ln|(x+1)^2/(1-x^2)|-ln|(x^2+1)/(1-x^2)|]+C，其中C是任意常数
2、令t=x-1，dx=dt
原式=∫(t^2+2t+2)/(t^2+1)^2dt
=∫[1/(t^2+1)+2t/(t^2+1)^2+1/(t^2+1)^2]dt
=arctant-1/(t^2+1)+∫dt/(t^2+1)^2
令t=tanu，dt=sec^2udu
原式=arctan(x-1)-1/(x^2-2x+2)+∫sec^2udu/sec^4u
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*∫2cos^2udu
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*∫(1+cos2u)du
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*[u+(1/2)*sin2u]+C
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*arctant+(1/2)*t/(t^2+1)+C
=(3/2)*arctan(x-1)+(x-3)/2(x^2-2x+2)+C，其中C是任意常数
原式=2*∫(tant+1)/[(tant-1)sec^4t]*sec^2tdt
=2*∫(sint+cost)/(sint-cost)*cos^2tdt
=∫(sint+cost)^2/(sin^2t-cos^2t)*(1+cos2t)dt
=∫(1+sin2t)/(-cos2t)*(1+cos2t)dt
=-∫(1+sin2t)/cos2tdt-∫(1+sin2t)dt
=-∫sec2tdt-∫tan2tdt-∫(1+sin2t)dt
=-(1/2)*ln|sec2t+tan2t|-(1/2)*ln|sec2t|-t+(1/2)*cos2t+C
=(1/2)*[cos2t-2t-ln|sec2t+tan2t|-ln|sec2t|]+C
=(1/2)*[(1-x^2)/(x^2+1)-2arctanx-ln|(x+1)^2/(1-x^2)|-ln|(x^2+1)/(1-x^2)|]+C，其中C是任意常数
2、令t=x-1，dx=dt
原式=∫(t^2+2t+2)/(t^2+1)^2dt
=∫[1/(t^2+1)+2t/(t^2+1)^2+1/(t^2+1)^2]dt
=arctant-1/(t^2+1)+∫dt/(t^2+1)^2
令t=tanu，dt=sec^2udu
原式=arctan(x-1)-1/(x^2-2x+2)+∫sec^2udu/sec^4u
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*∫2cos^2udu
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*∫(1+cos2u)du
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*[u+(1/2)*sin2u]+C
=arctan(x-1)-1/(x^2-2x+2)+(1/2)*arctant+(1/2)*t/(t^2+1)+C
=(3/2)*arctan(x-1)+(x-3)/2(x^2-2x+2)+C，其中C是任意常数
全部回答
- 1楼网友：雾月
- 2021-01-30 14:51
- 2楼网友：往事隔山水
- 2021-01-30 14:28
都大学了，还来百度问题。。
我要举报
如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
点此我要举报以上问答信息
大家都在看
推荐资讯